Question

A toy train rolls around a horizontal 1.0-m-diameter track. the coefficient of rolling friction is 0.10. how long does it take the train to stop if it's released with an angular speed of 40 rpm?

Answer 1

Deceleration due to rolling resistance (α) = coefficient*g/R = 0.1*9.81/0.5 = 1.962 rad/sec^2

Equations of motion:
ωf=ωo+αt
But ωf = 0 rad/sec; ωo= 2*pi*RPM/60 = 4.19 rad/sec

Therefore,
0=4.19-1.962t => 1.962t = 4.19 => t = 4.19/1.962 = 2.13 seconds

Answer 2

The train with a constant angular deceleration will stop in $\fbox{2.13\text{ s}}$.  

Further Explanation:

An object in pure rolling motion on a rough surface experiences resistive rolling force.

Given:

The diameter of wheel is $1\,{\text{m}}$.

The angular speed is $40\,{\text{rpm}}$.

Concept:

The equation of resistive or friction force is:

$\fbox{\begin\\F_r=m\cdot\text{a}_r\end{minispace}}$

Here, $m$ is the mass of body, ${a_r}$ is the deceleration due to rolling resistance and ${F_r}$ is the frictional force.

The equation of normal force is:

$\fbox{N = mg}$

Here, $g$ is the gravitational acceleration, $m$ is the mass of body, and $N$ is the normal force (which is equal to weight of body).

For body to be in equilibrium the equation of forces is:

${F_r} = \dfrac{{bN}}{R}$

 

Here, $b$ is the coefficient of rolling friction and $R$ is the radius of wheel.

Substitute $m{a_r}$ for ${F_r}$ and $mg$ for $N$ in above equation.

$\begin{gathered}m{a_r}=\frac{{bmg}}{R}\\{a_r}=\frac{{bg}}{R}\\\end{gathered}$

Here, ${a_r}$ is the deceleration due to rolling resistance, $b$ is the coefficient of rolling friction, $g$ is the acceleration due to gravity and $R$ is the radius of wheel.

Substitute $0.1$ for $b$, $9.81\text{ m}/\text{s}^2$ for $g$, and $\dfrac{1}{2}$ for $R$ in above equation.

$\begin{gathered}{a_r}=\frac{{0.1\times9.81\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}}{{0.5}}\\=1.962\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\\end{gathered}$

The initial angular velocity of body is:

${\omega _0} = \dfrac{{2\pi N}}{{60}}$

 

Here, $N$ is the angular speed in rpm and ${\omega _0}$ is the initial angular velocity.

Substitute $40\,{\text{rpm}}$ for $N$ in above equation.

$\begin{gathered}{\omega _0}=\frac{{2\pi\times40\,{\text{rpm}}}}{{60}}\\=4.19\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{gathered}$

Applying equation of motion:

${\omega _f}={\omega _0}-{\alpha _r}t$

Rearrange the above equation for value t:

$t=\dfrac{{{\omega _0}-{\omega _f}}}{{{\alpha _r}}}$

Here, ${\omega _f}$ is the final velocity, ${\omega _0}$ is the initial velocity, ${\alpha _r}$ is the deceleration and $t$ is the time taken by train to come to rest.

Substitute $0\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ for ${\omega _f}$ and $1.962\,{{{\text{rad}}} \mathord{\left/ {\vphantom {{{\text{rad}}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}$ for ${\alpha _r}$ and $4.19\,{{{\text{rad}}} \mathord{\left/ {\vphantom {{{\text{rad}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${\omega _0}$ in above equation.

$\begin{aligned}\\t&=\frac{{4.19\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}-0}}{{1.962\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}}\\&=2.13\,{\text{s}}\\\end{aligned}$

 

Thus, the train with a constant angular deceleration will stop in $\fbox{2.13\,{\text{s}}}$.

Learn more:

1.  Motion of a block under friction brainly.com/question/7031524

2.  Conservation of momentum in collision brainly.com/question/9484203

3. Motion of a ball under gravity brainly.com/question/10934170

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

Horizontal track, angular speed, coefficient of rolling friction, 40 rpm, and deceleration.